Question

How do you find the derivative of:

ln(3x^2 + x - 1)

I know the derivative of lnx is 1/x

Answer 1

If y = ln(junk)

then dy/dx = junk'/junk

so y' = (6x+1)/(3x^2 + x - 1)

Answer 2

[AI-generated answer — Explain Bot]

To find the derivative of ln(3x^2 + x - 1), you can use the chain rule. The chain rule states that if you have a function f(g(x)), the derivative of that function with respect to x is equal to the derivative of the outer function f'(g(x)) multiplied by the derivative of the inner function g'(x).

In this case, the outer function is ln(x) and the inner function is 3x^2 + x - 1. To find the derivative, you need to find the derivative of the outer function and the derivative of the inner function.

The derivative of ln(x) is 1/x, as you mentioned. So, the derivative of the outer function is 1/(3x^2 + x - 1).

Next, you need to find the derivative of the inner function. The inner function is a polynomial, so you can use the power rule. The power rule states that if you have a term of the form ax^n, the derivative with respect to x is nax^(n-1).

In this case, the derivative of 3x^2 is 6x, the derivative of x is 1, and the derivative of -1 (a constant) is 0.

Finally, you can multiply the derivatives of the outer and inner functions together:

(1/(3x^2 + x - 1)) * (6x + 1 + 0)

Simplifying this expression will give you the final result.